Calculate the weight of each of the following masses on earth. Show working

Suppose the half-life of an element is 10 years. How many half-lives will it take before only about 6% of the original sample re

sergiy2304 [10]

You don't need to worry about the 10 year bit with this question. Just grab a calculator and divide 100/2, then the answer to that (50) by 2 etc and keep dividing by 2 until you get down to 6.25.

The answer ends up being 4 half lives :)

If you don't understand what a half life is please let me know :)

4 0

3 years ago

Read 2 more answers

Which of the following is an effect of the electric force?

stich3 [128]

As we know that two charges exert force on each other when they are placed near to each other

The force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1, q_2$ = two different point charges

r = distance between two point charges

also we know that two similar charges always repel each other while two opposite charges always attract each other

so here correct answer would be

<em>A. A positive and negative charge attract each other.</em>

6 0

2 years ago

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A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary

Snezhnost [94]

Explanation:

Let $N_p\ and\ N_s$ are the number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p}{N_s}=\dfrac{1}{3}$

A resistor R connected to the secondary dissipates a power $P_s=100\ W$

For a transformer, $\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}$

$V_s=(\dfrac{N_s}{N_p})V_p$

$V_s=3V_p$ ...............(1)

The power dissipated through the secondary coil is :

$P_s=\dfrac{V_s^2}{R}$

$100\ W=\dfrac{V_s^2}{R}$

$V_p^2=\dfrac{100R}{9}$ .............(2)

Let $N_p'\ and\ N_s'$ are the new number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p'}{N_s'}=\dfrac{1}{24}$

New voltage is :

$V_s'=(\dfrac{N_s'}{N_p'})V_p'$

$V_s'=24V_p$ ...............(3)

So, new power dissipated is $P_s'$

$P_s'=\dfrac{V_s'^2}{R}$

$P_s'=\dfrac{(24V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}$

$P_s'=6400\ Watts$

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0

3 years ago

A mixture of helium and oxygen is used in scuba diving tanks to help prevent ""the bends"". 46 L helium and 12 L oxygen are comb

marin [14]

Answer

given,

For helium

Volume,V = 46 L

Pressure,P = 1 atm

Temperature,T = 25°C = 273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₁ = ?

number of moles

we know

P V = n R T

$n_1 =\dfrac{46 \times 1}{0.0821\times 298}$

n₁ = 1.89 moles

For oxygen

Volume,V = 12 L

Pressure,P = 1 atm

Temperature,T = 25°C = 273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₂ = ?

number of moles

we know

P V = n R T

$n_2 =\dfrac{12 \times 1}{0.0821\times 298}$

n₂ = 0.49 moles

Total volume of tank = 5 L

temperature of tank = 298 K

Partial pressure of helium

$P_1=\dfrac{n_1 R T}{V}$

$P_1=\dfrac{1.89\times 0.0821\times 298}{5}$

P₁ = 9.25 atm

Partial pressure of oxygen

$P_2=\dfrac{n_2 R T}{V}$

$P_2=\dfrac{0.49\times 0.0821\times 298}{5}$

P₂ = 2.39 atm

total pressure

P = P₁ + P₂

P = 9.25 + 2.39

P = 11.64 atm

8 0

2 years ago

Cuando una persona sube y baja una escalera, Cuanto vale su desplazamiento y cual es la medida de su trayectoria.

adoni [48]

Answer:

Primero, definimos el desplazamiento como la distancia entre la posición final y la posición inicial.

Así, si comenzamos abajo, luego subimos la escalera, y luego bajamos, la posición final y la posición inicial serán la misma

por lo que el desplazamiento es igual a cero.

La medida recorrida es el espacio total recorrido.

Es decir, si entre el principio y el final de la escalera hay una distancia D.

La persona que sube y baja, recorre esta distancia dos veces.

Entonces cuando una persona sube y baja la escalera, la medida de su trayectoria será 2*D.

8 0

3 years ago